Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Chemistry
The CFSE for [.CoCl6].4 - complex is 18000 cm- 1 . The Δ for [.CoCl4].2 - will be
Q. The CFSE for
[
C
o
C
l
6
]
4
−
complex is 18000
c
m
−
1
. The
Δ
for
[
C
o
C
l
4
]
2
−
will be
2605
192
NTA Abhyas
NTA Abhyas 2020
Coordination Compounds
Report Error
A
9000
c
m
−
1
43%
B
4000
c
m
−
1
11%
C
8000
c
m
−
1
43%
D
2000
c
m
−
1
3%
Solution:
The relation of CFSE for tetrahedral and octahedral complex is given as
Δ
t
=
9
4
Δ
0
So
Δ
t
for
[
C
o
C
l
4
]
2
−
=
9
4
×
18000
c
m
−
1
=
8000
c
m
−
1
.